A swiss cheese theorem for linear operators with two invariant subspaces

Abstract:

We study systems $(V,T,U_1,U_2)$ consisting of a finite dimensional vector space $V$, a nilpotent $k$-linear operator $T:V\to V$ and two $T$-invariant subspaces $U_1\subset U_2\subset V$. Let $\mathcal S(n)$ be the category of such systems where the operator $T$ acts with nilpotency index at most $n$. We determine the dimension types $(\dim U_1, \dim U_2/U_1, \dim V/U_2)$ of indecomposable systems in $\mathcal S(n)$ for $n\leq 4$. It turns out that in the case where $n=4$ there are infinitely many such triples $(x,y,z)$, they all lie in the cylinder given by $|x-y|,|y-z|$, $|z-x|\leq 4$. But not each dimension type in the cylinder can be realized by an indecomposable system. In particular, there are holes in the cylinder. Namely, no triple in $(x,y,z)\in (3,1,3)+\mathbb N(2,2,2)$ can be realized, while each neighbor $(x\pm 1,y,z), (x,y\pm 1,z),(x,y,z\pm 1)$ can. Compare this with Bongartz’ No-Gap Theorem, which states that for an associative algebra $A$ over an algebraically closed field, there is no gap in the lengths of the indecomposable $A$-modules of finite dimension.